Question

the two half - reactions are given below.
$2e^- + 2h_2o + hpo_3^{2-} \
ightarrow h_2po_2^- + 3oh^-$
$4oh^- + n_2h_4 \
ightarrow n_2 + 4h_2o + 4e^-$
what is the coefficient for the hydroxide ion in the combined balanced net ionic equation?

Explanation:

Step1: Balance electrons in both half - reactions

The first half - reaction: \(2e^-+2H_2O + HPO_3^{2 - }
ightarrow H_2PO_2^-+3OH^-\) (electrons gained = 2)
The second half - reaction: \(4OH^-+N_2H_4
ightarrow N_2 + 4H_2O+4e^-\) (electrons lost = 4)

To balance the number of electrons, we multiply the first half - reaction by 2.

Step2: Multiply the first half - reaction by 2

After multiplying the first half - reaction by 2, we get:
\(4e^- + 4H_2O+2HPO_3^{2 - }
ightarrow 2H_2PO_2^-+6OH^-\)

Step3: Add the two half - reactions

Now add the modified first half - reaction and the second half - reaction:

\((4e^- + 4H_2O+2HPO_3^{2 - })+(4OH^-+N_2H_4)=(2H_2PO_2^-+6OH^-)+(N_2 + 4H_2O+4e^-)\)

Cancel out the common terms ( \(4e^-\), \(4H_2O\) from both sides):

\(2HPO_3^{2 - }+N_2H_4 + 4OH^-=2H_2PO_2^-+N_2+6OH^-\)

Now, simplify the \(OH^-\) terms. Subtract \(4OH^-\) from both sides:

\(2HPO_3^{2 - }+N_2H_4=2H_2PO_2^-+N_2 + 2OH^-\)

Answer:

The coefficient of \(OH^-\) (hydroxide ion) in the balanced net ionic equation is 2.